Draw the molecules IFs and PF5. How do the molecular shapes differ? Оа
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In IFs there is a lone pair of electrons on iodine where as in PFs there are no lone pairs on phosphorus. As a result, the molecular
shape of IFs is square pyramidal and PFs is trigonal bipyramidal.
In IFs there is a lone pair of electrons on iodine where as in PFs there are no lone pairs on phosphorus. As a result, the molecular
shape of IFs is octahedral and PF5 is trigonal bipyramidal.
In PFs there is a lone pair of electrons on phosphorus where as in IFs there are no lone pairs on iodine. As a result, the molecular
shape of PFs is square pyramidal and IFs is trigonal bipyramidal.
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For questions 1 and 2, consider the following experimental data.hydrogen emission lines were detected at the following wavelengths (in nm): 121.6102.697.395.093.8question 1use the electromagnetic radiation classifications below and figure 1-1 in the introductory information for this lab (in the lab manual) to determine the nf value for the experimental data provided? wavelength, ? (nm) 650 700 550 600 400 450 500 visible spectrum wavelength, ? (m) 11 10 3 10 10 10 8 10 5 10 10 -10 10 9 10 10 10 10 -12 10 microwave radio infrared x-ray ultraviolet gamma 1020 1019 1018 1 1016 015 1014 01 12 109108 frequency, v (hz)a.1b. 2c. 3d. 4e. 5question 2using the data for the emission line with the longest wavelength, the known value of nf (from question 1 in this prelab), and the value of ni (deduced from the ? and nf values) calculate the rydberg constant for hydrogen (rh) in units of m-1.a) 1.097 x 10-11 m-1b) 5.921 x 107 m-1c) 1.097 x 10-2 m-1d) 9.252 x 106 m-1e) 1.097 x 107 m-1